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What does the math notation $\sum$ mean?

My school's prescribed book uses the weird letter E character without explaining what it is in the first chapter when it talks about the binomial equation. I can't find it on Google either because I don't know what it means or its name. Please help me!

$$ (x+a)^n = \sum_{k=0}^{n} \binom{n}{k}x^ka^{n-k}$$

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marked as duplicate by t.b., robjohn, Srivatsan, J. M., Asaf Karagila Dec 23 '11 at 14:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

It's the Greek capital letter $\Sigma$ sigma. Roughly equivalent to our 'S'. It stands for 'sum'. Read this for starters. – Jyrki Lahtonen Nov 14 '11 at 7:27 – pedja Nov 14 '11 at 7:28
...and maybe it wouldn't hurt to mention that sigma is a consonant, since called it a "weird E letter" could create a different impression. – Michael Hardy Nov 14 '11 at 12:04
In general $\displaystyle{\sum_{n=i}^{j} f(n) = f(i) + f(i+1) + \ldots + f(j-1) + f(j)}$ – badp Nov 14 '11 at 14:12
up vote 21 down vote accepted

This is a capital sigma. Its use is best illustrated by an example:

$$ \sum_{k = 1}^4 \frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}. $$

You begin by replacing the index (in this case, $k$) with the first value it takes on (in this case, 1). You then proceed to the next number and keep doing this replacement until you are at the upper limit (in this case, 4). Finally, you add all these terms up.

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As explained by Austin Mohr, it is the summation operation $\sum\limits_{k=m}^n f(k)$ (Greek uppercase letter Sigma) which sums every value of $f(k)$ where every value between $m$ and $n$ inclusively is substituted into $k$. That is:

$$\sum_{k=1}^5\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \approx{2.28}$$

A related operation is the product operator denoted $\prod\limits_{k=m}^n f(k)$ (Greek uppercase letter Pi) which returns the product of the terms with $k$ substituted for every value between $m$ and $n$ inclusive.

$$\prod_{k=1}^3 (k + x) = (1 + x) (2 + x) (3 + x) = x^3 + 5x^2 + 11x + 6$$

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